﻿ IIT - JEE Main Important Questions of Relations and Functions | Zigya

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# Relations and Functions

#### Multiple Choice Questions

1.

A real valued function f(x) satisfies the functional equation f(x – y) = f(x) f(y) – f(a – x) f(a + y) where a is a given constant and f(0) = 1, f(2a – x) is equal to

• –f(x)

• f(x)

• f(a) + f(a – x)

• f(-x)

A.

–f(x)

f(a – (x – a)) = f(a) f(x – a) – f(0) f(x)

= - f(x) [ ∵ x = 0, y= 0, f(0) = f2 (0)-f2(a) = 0 ⇒ f(a) = 0]

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2.

Suppose f(x) is differentiable x = 1 and

• 3

• 4

• 5

• 6

C.

5

As the function is differentiable so it is continuous as it is given that  and hence f(1) = 0 Hence
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3.

Let f and g be differentiable functions satisfying g′(a) = 2, g(a) = b and fog = I (identity function). Then f ′(b) is equal to

• 1/2

• 2

• 2/3

• None of these

A.

1/2

Given fog = I
⇒ fog(x) = x for all x
⇒ f ′(g(x)) g′(x) = 1 for all x

f ′(g(a)) =1/g('a) = 1/2
⇒f(b') = 1/2

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4.

The range of the function 7-xPx-3 is

• {1, 2, 3}

• {1, 2, 3, 4, 5}

• {1, 2, 3, 4}

• {1, 2, 3, 4, 5, 6}

A.

{1, 2, 3}

The given function f(x) = 7-xPx-3 would be defined if
(i) 7 - x > 0 ⇒ x < 7
(ii) x - 3 > 0 ⇒ x > 3
(iii) (x - 3) < (7 - x)
⇒ 2x < 10 ⇒ x < 5
⇒ x = 3, 4, 5
Hence Range of f(x) = {4P0, 3P1, 2P2}
Range of f(x) = {1, 3, 2}

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5.

Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)} be a relation on the set A = {3, 6, 9, 12} be a relation on the set A = {3, 6, 9, 12}. The relation is

• reflexive and transitive only

• reflexive only

• an equivalence relation

• reflexive and symmetric only

A.

reflexive and transitive only

Reflexive and transitive only.
e.g. (3, 3), (6, 6), (9, 9), (12, 12) [Reflexive]
(3, 6), (6, 12), (3, 12) [Transitive].

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6.

Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relation on the set A = {1, 2, 3, 4}. The relation R is

• a function

• reflexive

• not symmetric

• transitive

C.

not symmetric

Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} is a relation on the set A = {1, 2, 3, 4}, then
(a) Since ∈ R and (2,3) ∈ R, so R is not a function.
(b) Since (1, 3) ∈ R and (3, 1) ∈ R but (1, 1) ∉ R. So R is not transitive.
(c) Since (2, 3) ∈ R but (3, 2) ∉ R, so R is not symmetric.
(d) Since (4, 4) ∉ R so R is not reflexive,

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7.

A function is matched below against an interval where it is supposed to be increasing. Which of the following pairs is incorrectly matched?

•  Interval Function (-∞, ∞) x3 – 3x2 + 3x + 3
•  Interval Function [2, ∞) 2x3 – 3x2 – 12x + 6
•  Interval Function (-∞, 1/3] 3x2 – 2x + 1
•  Interval Function (- ∞, -4] x3 + 6x2 + 6

C.

 Interval Function (-∞, 1/3] 3x2 – 2x + 1

Clearly function f(x) = 3x2 – 2x + 1 is increasing when
f′(x) = 6x – 2 ≥ 0 ⇒ x∈[1/3, ∞)

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8.

Let f : (-1, 1) → B, be a function defined by  then f is both one-one and onto when B is the interval

• [0, π/2)

C.

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9.

The domain of the function

• [2, 3]

• [2, 3)

• [1, 2]

• [1, 2)

B.

[2, 3)

For the the function
we will define

(I) −1≤ (x − 3) ≤1⇒ 2 ≤ x ≤ 4 ....... (i)
(II) 9 x 0 3 x 4 2 − ≥ ⇒ − < < ....... (ii)
From relation (i) and (ii), we get 2 < x < 3
∴ Domain of the given function = [2, 3)

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10.

The graph of the function y = f(x) is symmetrical about the line x = 2, then

• f(x + 2)= f(x – 2)

• f(2 + x) = f(2 – x)

• f(x) = f(-x)

• f(x) = - f(-x)

B.

f(2 + x) = f(2 – x)

If y = f (x) is symmetric about the line x = 2 then f(2 + x) = f(2 – x).

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